# Topological retraction vs categorical retraction

Let $(X,\tau)$ be a topological space. We say that $A\subseteq X$ is a

1. topological retract if there is a continuous map $r:X\to A$ onto a subspace $A \subseteq X$ such that for all $a\in A$ we have $r(a) = a$;
2. categorical retract if there are continouus maps $r: X\to A$ and $f: A\to X$ such that $r \circ f = \textrm{id}_A$. (That is $r: X\to A$ is a retraction in the categorical sense. And obviously, $f$ has to be injective, but we do not require that $f$ is the inclusion map!)

Obviously, any topological retract is categorical. Is there an example of a categorical retract that is not a topological retract?

• In the second definition $i$ is supposed to be the inclusion of $A$ into $X$ (otherwise why mention the inclusion at all?). The two agree. – Qiaochu Yuan Dec 8 '14 at 10:00
• As Qiaochu says, this is a strange question, since the inclusion $A\subseteq X$ is manifest in the first condition but not the second. With the more generous reading, though, an example is given by any null-homotopic embedded circle in the torus. – Mark Grant Dec 8 '14 at 10:06
• I fixed the first definition so that it makes sense. – Andrej Bauer Dec 8 '14 at 10:09
• No $\iota: A\to X$ can be any continous map, but you are right in that it has to be injective -- but it doesn't have to be the injection! Maybe my choice of letter was unfortunate, I will change this. – Dominic van der Zypen Dec 8 '14 at 10:09
• It follows that $\iota$ is injective! – Andrej Bauer Dec 8 '14 at 10:09

The two notions agree.

Clearly, if we have a topological retraction $r : X \to A$ we may take $f : A \to X$ to be the subspace inclusion.

Conversely, given a categorical retraction $r : X \to A$, $f : A \to X$ we have the corresponding topological retraction $r' : X \to A'$ where $A' = \mathsf{im}(f) \subseteq X$ and $r' = f \circ r$. Observe that $f : A \to A'$ is a homeomorphism since it has a continuous inverse, namely $r$ restricted to $A'$: for every $x \in A'$ there is $y \in A$ such that $x = f(y)$ and so $$f(r(x)) = f(r(f(y)) = f(y) = x.$$ The other direction is obvious as $r \circ f = \mathsf{id}_A$ by assumption.

This argument is going to work in any category with well-behaved images.

• This only shows that $A'$ is a topological retract of $X$, but $A$ need not be. See Mark Grant's counterexample! – Marc Hoyois Dec 8 '14 at 13:38
• @MarcHoyois: I think Andrej's reading of the OP's definition of topological retract is "any space which is homeomorphic to a retract", and it seems that is what the OP intended. The fact that $A$ is given as a subset of $X$ is something of a red herring. – Mark Grant Dec 8 '14 at 14:14
• @MarcHoyois: that's completely irrelevant. Of course $A$ need not be a subspace of $X$, there's no way to make something into a subspace if it isn't (this is a case of set theory doing harm). The answer I give is the best possible: there is a canonical subspace $A'$ which is homeomorphic to $A$. Moreover, the pair of arrows $(r,f)$ is isomorphic to the pair of arrows $(f \circ r, \mathsf{incl}_{\mathsf{im}(f)})$ in the relevant category (whose objects are pairs of arrows $A \to X$ and $X \to A$ and morphisms are the evident ones). – Andrej Bauer Dec 8 '14 at 16:08

Yes. Take $X$ to be the standard torus and $A\subseteq X$ to be any null-homotopic embedded circle. Then $A$ is not a topological retract (since the inclusion fails to be injective on homology groups, say) but it is a categorical retract, since we can just take the map $f: A\to X$ which includes the circle as the first factor (edit: and $r: X\to A$ to be projection onto the first factor).

• This doesn't work as you have to provide $r : X \to A$ as well. – Andrej Bauer Dec 8 '14 at 10:16
• @Andrej: In my reading of the question, $A$ is regarded as a subset of $X$ in condition 1, but as an abstract topological space (with the subspace topology inherited from $X$) in condition 2. In fact I was answering the original question (before your edit), but I still don't see what's wrong with my answer. – Mark Grant Dec 8 '14 at 10:30
• @MarkGrant -- you have provided an illustration instead of a full answer. – Włodzimierz Holsztyński Dec 8 '14 at 11:05
• @Emil: It is not a retract, in the sense that there does not exist a map $r: X\to A$ which restricts to the identity on $A$. This can be seen by looking at first homology groups. – Mark Grant Dec 8 '14 at 12:01
• I see. Under that reading you indeed have a counter-example, but that reading is completely unreasonable. If that's what the OP asked, then he should be told that's the wrong thing to ask. – Andrej Bauer Dec 8 '14 at 16:10